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It is a job interview question.

So, what's the value of a vanilla European call option of infinite maturity, and a given strike, vol, interest rate, spot price.

I think, the answer should be "zero".

The contract never pays, because infinite maturity will never be reached.

It should not be equal to the spot price, which BS formula suggests in the limit T goes to infinity, I think.

If it were an American call with infinite maturity, the price could be anywhere between S and S-K.

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  • $\begingroup$ What are the purpose and payoff of a European option with infinity maturity? If the option will never be exercised, why do you need this option? $\endgroup$
    – Gordon
    May 27 '19 at 16:43
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    $\begingroup$ Given that the option will never be exercised, it has a zero value . $\endgroup$
    – Gordon
    May 27 '19 at 16:48
  • $\begingroup$ @Gordon this is interesting because the math suggested differently .... hmm $\endgroup$
    – Sanjay
    May 27 '19 at 18:55
  • $\begingroup$ @Sanjay: the math is based on a maturity and then take a limit. However, if we start with a infinity maturity, then the payoff is not well defined, given that you will never have the chance to exercise -- this is mentioned in my first question on the payoff. $\endgroup$
    – Gordon
    May 27 '19 at 20:04
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So we are assuming that the stock does not pay any dividend? Then it should be $S_0$, which is the limiting behaviour of the BS formula. $\Phi ( d_1)$ goes to 1 as maturity approaches infinity, and $\Phi (d_2)$ can only be between 0 and 1, so the $e^{-r\tau}$ will make the $K e^{-r\tau} \Phi (d_2)$ zero.

I can see the argument around the payoff at infinity not of any value, but then if the stock never pays dividend, then can’t you make the same argument about the stock payoff? So call option price equal to $S_0$ make sense.

For a dividend paying stock, it should be zero as you can easily verify by looking at the extended BS formula. The second term goes to zero as above, but the first term now has got the term $e^{-r_f \tau}$ so it goes to zero as well. Again this makes sense because the stock value will mainly be coming from the dividend stream, which the option lacks, so its worth nothing.

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I'm not sure there has ever existed a perpetual European call option, but I'm happy to indulge in the thought process. I say it can't be worth zero, because there are certain events that cause it to have value: a) if the option were subject to variation margin according to the market value, then obviously the market could decide the value is non zero. In this situation, theoretical arguments are moot- if you have sold the option, you will have to margin it, and if the other side of the trade has more liquidity than you, you will lose the battle. b) if there is no variation margin, you have a stronger argument- for one thing, who are you buying the option from? Given an infinite time frame, they will eventually go bankrupt. But there is still a potential for value: what if the underlying stock is subject to a corporate event such as a cash acquisition? Depending on the documentation, that could deliver cash to option holders. Even if the company goes bankrupt, there is sometimes residual value for stockholders.

So I say it cannot be zero, although you have a strong case that it is nowhere near the BS formula limit of S.

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In the standard BS model the price follows a GBM so the company can never be worth zero. But, in practice, the underlying can become worthless.

In the GBM case,the price equals $s_0$. But, in reality there is a positive probability that $S$ becomes worthless and stays that way forever. Therefore, $E[S_T]$ must tend to zero as $T$ tends to infinity and the option must be worthless. If we find an asset that truly can keep its value forever, then $s_0$ seems fair? An example: at the end of times you have the option to jump back in time to today. Even though the point will never be reached, it will still always have the same value as your life has today.

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